PK 2017 Q16
16. Seven positive integers \(a\), \(b\), \(c\), \(d\), \(e\), \(f\), \(g\) are written in a row. Every number differs by one from its neighbours. The total of the seven numbers is 2017. Which of the numbers can be equal to 286?
- A only \(a\) or \(g\)
- B only \(b\) or \(f\)
- C only \(c\) or \(e\)
- D only \(d\)
- E any of them
Pink Kangaroo 2017, Question 16. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: A only \(a\) or \(g\)
Each number \(a\), \(b\), \(c\), \(d\), \(e\), \(f\), \(g\) differs from its neighbour by one, so they alternate odd and even. To obtain an odd total, we must have an odd number of odd numbers in the list. Hence \(b\), \(d\), \(f\) are odd and cannot be equal to 286.
If \(c = 286\), then the biggest total possible is 288 + 287 + 286 + 287 + 288 + 289 + 290 = 2015 which is too small. By reversing this list, we can also rule out \(e = 286\).
We can obtain the total 2017 if we start with \(a = 286\) since 286 + 287 + 288 + 289 + 290 + 289 + 288 = 2017. By reversing this, we could also end with \(g = 286\).
Solution © UK Mathematics Trust, from the PK 2017 solutions.